X = number of times Roberto throws the baseball in the strike zone X can take on the following values: 0, 1, 2, 3
p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial) n = 3 = sample size = number of times the baseball is thrown
Binomial Probabilities:
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k) P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0) P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3) P(X = 0) = (1)*(0.72)^(0)*(0.28)^3 P(X = 0) = (1)*(1)*(0.021952) P(X = 0) = 0.021952 P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k) P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1) P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2) P(X = 1) = (3)*(0.72)^(1)*(0.28)^2 P(X = 1) = (3)*(0.72)*(0.0784) P(X = 1) = 0.169344 P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k) P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2) P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1) P(X = 2) = (3)*(0.72)^(2)*(0.28)^1 P(X = 2) = (3)*(0.5184)*(0.28) P(X = 2) = 0.435456 P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k) P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3) P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0) P(X = 3) = (1)*(0.72)^(3)*(0.28)^0 P(X = 3) = (1)*(0.373248)*(1) P(X = 3) = 0.373248 P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)
The table will look like what you see in the attached image
----------------------------- Problem 1, part B
Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0
0.169+0.435+0.373 = 0.977
So the probability of at least one of the baseballs hits the strike zone is 0.977
========================================= Problem 2
Convert each raw x score to a z score
Company A z = (x-mu)/sigma z = (260-276)/5.8 z = -2.759 ---------------- Company B z = (x-mu)/sigma z = (260-252)/3.4 z = 2.353 ---------------- The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A
